A curve `y=f(x)`passesthrough point `P(1,1)`. The normal to the curve at `P`is a `(y-1)+(x-1)=0`. If the slope of the tangent at any point on thecurve is proportional to the ordinate of the point, then the equation of thecurve is(a)`( b ) (c) y=( d ) e^(( e ) (f) K(( g ) (h) x-1( i ))( j ))( k ) (l)`(m) (b) `( n ) (o) y=( p ) e^(( q ) (r) K e (s))( t ) (u)`(v)(c)`( d ) (e) y=( f ) e^(( g ) (h) K(( i ) (j) x-2( k ))( l ))( m ) (n)`(o) (d) None of these

A curve `y=f(x)`passesthrough point `P(1,1)`. The normal to the curve

1.	A curve `y=f(x)`passesthrough point `P(1,1)`. The normal to the curve at `P`is a `(y-1)+(x-1)=0`. If the slope of the tangent at any point on thecurve is proportional to the ordinate of the point, then the equation of thecurve is(a)`( b ) (c) y=( d ) e^(( e ) (f) K(( g ) (h) x-1( i ))( j ))( k ) (l)`(m) (b) `( n ) (o) y=( p ) e^(( q ) (r) K e (s))( t ) (u)`(v)(c)`( d ) (e) y=( f ) e^(( g ) (h) K(( i ) (j) x-2( k ))( l ))( m ) (n)`(o) (d) None of these
Answer» Correct Answer - `y = e^(a(x-1)), (1+(e^(-a))/(a) - (1)/(2a))`

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
The area bounded by the curves `y=cosx` and `y=sinx` between the ordinates `x=0` and `x=(3pi)/2` isA. `4sqrt(2) + 2`B. `4sqrt(2) +1`C. `4sqrt(2) + 1`D. `4sqrt(2) - 2`
Obtain the area enclosed by region bounded by the curves `y = x ln x` and ` y = 2x-2x^(2)`.A. `7//6`B. `7//24`C. `12//7`D. `7//12`
If `I_n=int_0^pi e^x(sinx)^n dx ,` then `(I_3)/(I_1)` is equal toA. `3//5`B. `1//5`C. `1`D. `2//5`
Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4
`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`